Question

2.Решите систему уравнений графическим способом: {x+y=5 {2x+y=63.Найдите:a) область определения функции(промежуток для переменной х), заданной формулой:1) y=14-2x2) y=x/(x+2)​ Смотреть ответ karolinak1002 karolinak1002 2. решение: {у=5-х {2х+у=6 2х+5-х=6 х=1 у=5-1 у=4 Новые вопросы в Алгебра Дві сім'ї вирушили на дитяче свято. Перша сім'я купила два дитячих квитки і один дорослий, і всього заплатила 81 гривень. Друга сім'я купила три дитяч … их квитки і два дорослих, і всього заплатила 142 гривень. Скільки коштує один дитячий квиток і один дорослий квиток? срочно!!!!! Упростите: 4a(a+2b)-7b(2a-b)-3a(2a-3b).​ Упростите : а) (14а^2-17ab-8b^2)-(17ab+12a^2-4b^2), б) 4а(а+2b)-7b(2a-b)-3a(2a-3b).​ Выполните действия: a) (4x-7)(2-3x), b) (1-2x)(5+x-x^2 ),в) (24x^4-12x^3-18x^2) : (-6x^2). ​ Розв'яжіть рівняння 2х4 - 5x2 + 2 = 0 срочно ​ Предыдущий Следующий Вместе мы знаем больше Этот сайт использует cookies Политика Cookies . 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Answer 1

Давайте сначала решим систему уравнений графическим способом:

Система уравнений:

1. x + y = 5
2. 2x + y = 6

Для этого мы можем нарисовать графики обоих уравнений на одном графике и найти точку их пересечения, которая будет решением системы.

Первое уравнение (x + y = 5) можно переписать в виде: y = 5 - x. Это уравнение представляет собой прямую линию с угловым коэффициентом -1 и смещением (свободным членом) 5 на оси y.

Второе уравнение (2x + y = 6) можно переписать в виде: y = 6 - 2x. Это также уравнение прямой с угловым коэффициентом -2 и смещением 6 на оси y.

Теперь нарисуем обе эти прямые на графике и найдем точку их пересечения:

scssCopy code
   |
6  *             + (2x + y = 6)
   |
5  *  *          +
   |     *       |
   |        *    |
   |           * |
   +------------------
      1    2    3    4  (x-ось)

Точка пересечения прямых находится при x = 1 и y = 4. Это и есть решение системы уравнений.

Теперь найдем область определения функций:

1. y = 14 - 2x Область определения (промежуток для переменной x) не ограничена, и x может принимать любые значения. Таким образом, область определения для этой функции - все действительные числа.

2. y = x / (x + 2) Область определения для этой функции определяется условием деления на ноль. В знаменателе (x + 2) не должно быть нулем, поэтому x + 2 ≠ 0. Решив это уравнение, получаем x ≠ -2. Таким образом, область определения для этой функции - все действительные числа, кроме x = -2.